题意：
给出长度为 n 的数组 a 和 q 个操作。
- 询问区间 [l, r] 内的元素之和
- 将 [l, r] 内的元素变为原来的三次方
对于第一个操作输出一行一个整数。
题解：
打表发现循环节是48，可以维护48个线段树，记录偏移量。
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 const int N = 1e5 + 10; 5 typedef long long ll; 6 const ll mod = 95542721; 7 inline ll read() 8 { 9 char ch = getchar();ll x = 0, f = 1; 10 while(ch < '0' || ch > '9') {if(ch == '-') f = -1; ch = getchar();} 11 while(ch >= '0' && ch <= '9') {x = (x << 1) + (x << 3) - '0' + ch; ch = getchar();} 12 return x * f; 13 } 14 ll n, m; 15 ll sum[N << 2][49]; 16 int lazy[N << 2]; 17 int tr[N << 2]; 18 19 20 void build(int rt, int l, int r) 21 { 22 if(l == r) 23 { 24 scanf("%lld", &sum[rt][0]); 25 for (int i = 1; i <= 47; i ++) 26 sum[rt][i] = sum[rt][i - 1] * sum[rt][i - 1] % mod * sum[rt][i - 1] % mod; 27 return ; 28 } 29 int mid = l + r >> 1; 30 build(rt << 1, l, mid); 31 build(rt << 1 | 1, mid + 1, r); 32 for (int i = 0; i <= 47; i ++) 33 sum[rt][i] = sum[rt << 1][i] + sum[rt << 1 | 1][i]; 34 } 35 36 void pushdown(int rt) 37 { 38 if(lazy[rt]) 39 { 40 lazy[rt << 1] = (lazy[rt << 1] + lazy[rt]) % 48; 41 lazy[rt << 1 | 1] = (lazy[rt] + lazy[rt << 1 | 1]) % 48; 42 tr[rt << 1] = (tr[rt << 1] + lazy[rt]) % 48; 43 tr[rt << 1 | 1] = (tr[rt << 1 | 1] + lazy[rt]) % 48; 44 lazy[rt] = 0; 45 } 46 } 47 48 49 void update(int rt, int L, int R, int l, int r) 50 { 51 if(L <= l && R >= r) 52 { 53 lazy[rt] = (lazy[rt] + 1) % 48; 54 tr[rt] = (tr[rt] + 1) % 48; 55 return ; 56 } 57 pushdown(rt); 58 int mid = l + r >> 1; 59 if(L <= mid) update(rt << 1, L, R, l, mid); 60 if(R > mid) update(rt << 1 | 1, L, R, mid + 1, r); 61 for (int i = 0; i <= 47; i ++) 62 sum[rt][i] = sum[rt << 1][(i + tr[rt << 1])%48] + sum[rt << 1 | 1][(i + tr[rt << 1 | 1]) % 48]; 63 tr[rt] = 0; 64 } 65 66 ll query(int rt, int L, int R, int l, int r) 67 { 68 if(L <= l && R >= r) 69 { 70 return sum[rt][tr[rt]]; 71 } 72 int mid = l + r >> 1; 73 pushdown(rt); 74 ll ans = 0; 75 if(L <= mid) ans = (ans + query(rt << 1, L, R, l, mid)) % mod; 76 if(R > mid) ans = (ans + query(rt << 1 | 1, L, R, mid + 1, r)) % mod; 77 return ans; 78 } 79 80 int main() 81 { 82 n = read(); 83 build(1, 1, n); 84 m = read(); 85 while(m --) 86 { 87 int op = read(), l = read(), r = read(); 88 if(op == 1) 89 { 90 printf("%lld ", query(1, l, r, 1, n)); 91 } 92 else 93 { 94 update(1, l, r, 1, n); 95 } 96 } 97 }